Answer to Question #260850 in Linear Algebra for talie

Question #260850

Show that the map T : R^


2 −→ R^2 defined by T(x1, x2) = (x1 −x2, 0) is a linear transformation, hence find the matrix


of this transformation relative to the basis B = {(−1, 1),(0, 1)}

1
Expert's answer
2021-11-15T19:52:28-0500

Solution;

Given ;

T(x1,x2)=[x1x20]T(x_1,x_2)=\begin{bmatrix} x_1-x_2\\ 0 \end{bmatrix}

A map T is linear if;

T(x)=AxT(x)=Ax

A is the matrix of transformation;

Here;

A=[1100]A=\begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix}

Clearly;

T(x1,x2)=[1100]T(x_1,x_2)=\begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix} [x1x2]\begin{bmatrix} x_1\\ x_2 \end{bmatrix} =(x1x2,0)(x_1-x_2,0)

Therefore the map is a linear transformation.

And;

B=[(1,1),(0,1)]B=[(-1,1),(0,1)]

Hence;

TB(1,1)=[110]=[20]T_B(-1,1)=\begin{bmatrix} -1-1\\ 0 \end{bmatrix}=\begin{bmatrix} -2\\ 0 \end{bmatrix}

TB(0,1)=[010]=[10]T_B(0,1)=\begin{bmatrix} 0-1\\ 0 \end{bmatrix}=\begin{bmatrix} -1\\ 0 \end{bmatrix}

Hence;

TB=T[(1,1),(0,1)]=[(2,0),(1,0)]T_B=T[(-1,1),(0,1)]=[(-2,0),(-1,0)]

Let;

(2,0)=a1(1,1)+b1(0,1)(-2,0)=a_1(-1,1)+b_1(0,1)

(2,0)=(a1,a1)+(0,b1)(-2,0)=(-a_1,a_1)+(0,b_1)

(2,0)=(a1,a1+b1)(-2,0)=(-a_1,a_1+b_1)

From which;

2=a1-2=-a_1 a1=2\therefore a_1=2

Also;

O=a1+b1O=a_1+b_1 ;

b1=a1=2b_1=-a_1=-2

Now,let;

(1,0)=a2(1,1)+b2(0,1)(-1,0)=a_2(-1,1)+b_2(0,1)

(1,0)=(a2,a2)+(0,b2)(-1,0)=(-a_2,a_2)+(0,b_2)

(1,0)=(a2,a2+b2)(-1,0)=(-a_2,a_2+b_2)

From which;

1=a2,a2=1-1=-a_2 ,\therefore a_2=1

Also;

0=a2+b20=a_2+b_2 ;

b2=a2=1b_2=-a_2=-1

The coefficient matrix A,is ;

A=[2211]A=\begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix}

Hence,the matrix of transformation of T relative to B is AT;

AT=[2121]A^T=\begin{bmatrix} 2 & 1 \\ -2 & -1 \end{bmatrix}





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